If A and B are non-singular matrices of order n, then
[where \( A^T \) denotes transpose of A]
Match List-I with List-II
| List-I | List-II |
| (A) \( \mathrm{adj}(AB) \) | (I) \( |A|^{n-1} \) |
| (B) \( \mathrm{adj}(\mathrm{adj}\,A) \) | (II) \( (\mathrm{adj}\,B)\ (\mathrm{adj}\,A) \) |
| (C) \( \mathrm{adj}(A^T) \) | (III) \( (\mathrm{adj}\,A)^T \) |
| (D) \( |\mathrm{adj}\,A| \) | (IV) \( |A|^{n-2}A \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
(A) \( \mathrm{adj}(AB) \)
This matches (II) \( (\mathrm{adj}\,B)(\mathrm{adj}\,A) \). The reversal law holds for adjoints as for inverses.
(B) \( \mathrm{adj}(\mathrm{adj}\,A) \)
This matches (IV) \( |A|^{n-2}A \). This is the standard result for the adjoint of the adjoint.
(C) \( \mathrm{adj}(A^T) \)
This matches (III) \( (\mathrm{adj}\,A)^T \). Taking the transpose and the adjoint can be done in either order.
(D) \( |\mathrm{adj}\,A| \)
This matches (I) \( |A|^{n-1} \). Take determinants in \( A\,(\mathrm{adj}\,A)=|A|\,I \): \( |A|\,|\mathrm{adj}\,A|=|A|^n \).
So the matching is (A) - (II), (B) - (IV), (C) - (III), (D) - (I).